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Basic Geometry : Plane Geometry

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Example Questions

Example Question #1141 : Plane Geometry

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

72.59

74.09

69.39

61.80

Correct answer:

61.80

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{14\sqrt2}{2}$

$\text{leg}=7\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=14+14+14+7\sqrt2+7\sqrt2$

$\text{Perimeter}=61.80$

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Example Question #1142 : Basic Geometry

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

48.20

69.59

70.63

80.05

Correct answer:

70.63

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{16\sqrt2}{2}$

$\text{leg}=8\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=16+16+16+8\sqrt2+8\sqrt2$

$\text{Perimeter}=70.63$

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Example Question #1142 : Plane Geometry

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

77.65

79.46

84.19

81.20

Correct answer:

79.46

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{18\sqrt2}{2}$

$\text{leg}=9\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=18+18+18+9\sqrt2+9\sqrt2$

$\text{Perimeter}=79.46$

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Example Question #161 : 45/45/90 Right Isosceles Triangles

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

97.82

88.28

85.61

81.09

Correct answer:

88.28

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{20\sqrt2}{2}$

$\text{leg}=10\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=20+20+20+10\sqrt2+10\sqrt2$

$\text{Perimeter}=88.28$

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Example Question #162 : 45/45/90 Right Isosceles Triangles

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

97.11

99.42

102.58

95.10

Correct answer:

97.11

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{22\sqrt2}{2}$

$\text{leg}=11\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=22+22+22+11\sqrt2+11\sqrt2$

$\text{Perimeter}=97.11$

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Example Question #161 : 45/45/90 Right Isosceles Triangles

A right isosceles triangle is stacked on top of a square as shown in the figure. Find the perimeter of the compound shape.

Possible Answers:

102.42

105.94

107.28

101.09

Correct answer:

105.94

Explanation:

Notice that the hypotenuse of the right isosceles triangle is also the length of a side of the square.

First, we will need to find the length of the legs of the triangle by using the Pythagorean theorem.

$\text{leg}^2+\text{leg}^2=\text{side}^2$

$2(\text{leg})^2=\text{side}^2$

$\text{leg}^2=\frac{\text{side}^2}{2}$

$\text{leg}=\sqrt{\frac{\text{side}^2}{2}}=\frac{\text{side}\sqrt2}{2}$

Substitute in the given length of the hypotenuse to find the length of the leg of the triangle.

$\text{leg}=\frac{24\sqrt2}{2}$

$\text{leg}=12\sqrt2$

In order to find the perimeter, add up the lengths outlined in red. The perimeter includes the two legs of the triangle and three sides of the square.

Therefore:

$\text{Perimeter}=24+24+24+12\sqrt2+12\sqrt2$

$\text{Perimeter}=105.94$

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Example Question #1 : How To Find The Length Of The Side Of A 45/45/90 Right Isosceles Triangle

The perimeter of a 45-45-90 triangle is 100 inches. To the nearest tenth of an inch, what is the length of each leg?

Possible Answers:

$26.8\ inches$

$41.4\ inches$

$29.2\ inches$

$33.3\ inches$

$20\ inches$

Correct answer:

$29.2\ inches$

Explanation:

Let be the length of a leg; then the hypotenuse is $c=a\sqrt{2}$ , and the perimeter is

$P = a + a + a \sqrt{2} = (2 + \sqrt{2} ) a = 100$

Therefore,

$a = \frac{100}{2 + \sqrt{2}} \approx 29.2$

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Example Question #1141 : Basic Geometry

$\fn_cm In\; \Delta ABC, \overline{BC}=3 \;and \;\angle BAC=45^{\circ}. \;What\;is\;the\;length\;of\;\overline{AB}?$

Possible Answers:

Correct answer:

Explanation:

$\fn_cm \Delta ABC\;is\;an\;isosceles\;triangle. \;In\; an \;isosceles\;triangle, two\;sides\;and \;their \;opposite\;angles \;are\;equal.$

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Example Question #3 : How To Find The Length Of The Side Of A 45/45/90 Right Isosceles Triangle

$\fn_cm What\;is\;the\;length\;of\;\overline{DE} \;in\; \Delta DEF?$

Possible Answers:

$2\sqrt2$

$\sqrt2$

Correct answer:

Explanation:

$\fn_cm \Delta DEF \;is\;a\;45^{\circ}-45^{\circ}-90^{\circ}\;isoscles\;triangle, \;which\;have\;sides\;in\;a\;ratio\;of\;1:1:\sqrt2.$

$You\;could\;also\;use\;the\;Pythagorean\;Theorem, \;knowing\;that\;the\;two\;legs\;are\;equal.$

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Example Question #1141 : Plane Geometry

Angle in the triangle shown below is 45 degrees. Side has a length of 10. What is the length of side ?

Right_triangle_sides_and_points

Possible Answers:

$5\sqrt{2}$

$\small 20$

$\small \frac{5}{\sqrt{2}}$

$\small 5$

$\small 8$

Correct answer:

$5\sqrt{2}$

Explanation:

Since we know two of the three angles in this triangle, we can calculate the third, .

B=180-90-45=45

Therefore this is a 45/45/90 right triangle. Remember that 45/45/90 right triangles are have a leg:leg:hypotenuse ratio of 1:1: $\small \sqrt{2}$ .

We know the hypotenuse, $\small c$ , so we can quickly calculate the length of one of the legs, $\small a$ , by dividing by $\small \sqrt{2}$ :

$\small a=\frac{c}{\sqrt{2}}=\frac{10}{\sqrt{2}}$

To make this look like one of the answer choies, rationalize the denominator by muliplying the fraction by $\frac{\sqrt{2}}{\sqrt{2}}$ :

$\frac{10}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}}=\frac{10\cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}=\frac{10\sqrt{2}}{2}=5\sqrt{2}$

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Basic Geometry : Plane Geometry

Study concepts, example questions & explanations for Basic Geometry

All Basic Geometry Resources

Example Questions

Example Question #1141 : Plane Geometry

Example Question #1142 : Basic Geometry

Example Question #1142 : Plane Geometry

Example Question #161 : 45/45/90 Right Isosceles Triangles

Example Question #162 : 45/45/90 Right Isosceles Triangles

Example Question #161 : 45/45/90 Right Isosceles Triangles

Example Question #1 : How To Find The Length Of The Side Of A 45/45/90 Right Isosceles Triangle

Example Question #1141 : Basic Geometry

Example Question #3 : How To Find The Length Of The Side Of A 45/45/90 Right Isosceles Triangle

Example Question #1141 : Plane Geometry

All Basic Geometry Resources

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